Cloaking-Lorentz - Infoscience - EPFL
Approximate cloaking for the full wave equation via change of variables: the Drude-Lorentz model Hoai-Minh Nguyen∗ and Michael S. Vogelius† August 26, 2014
Abstract This paper concerns approximate cloaking by mapping for a full, but scalar wave equation, when one allows for physically relevant frequency dependence of the material properties of the cloak. The paper is a natural continuation of [20], but here we employ the Drude-Lorentz model in the cloaking layer, that is otherwise constructed by an approximate blow up transformation of the type introduced in [10]. The central mathematical problem translates into the analysis of the effect of a small inhomogeneity in the context of a non-local full wave equation.
1
Introduction
Cloaking by mapping (frequently referred to as transformation optics) was introduced by Pendry, Schurig, and Smith [23] for the Maxwell system, and Leonhardt [12] in the geometric optics setting. These authors used a singular change of variables which blows up a point to a cloaked region. The exact same transformation had been used before by Greenleaf, Lassas, and Uhlmann [6] to establish non-uniqueness in the context of the Calderon problem. The singular nature of the cloaks presents various difficulties in practice as well as in theory: (1) they are hard to fabricate and (2) in certain cases the correct definition of the corresponding electromagnetic fields is not obvious. To avoid the use of singular structures, regularized schemes have been proposed in [3, 4, 10, 26, 27]. In this paper we analyse approximate cloaking for a full wave equation using transformation optics, where we incorporate the Drude-Lorentz model, see e.g., [8], in the layer constructed by transformation optics. The Drude-Lorentz model takes into account the effect of the oscillations of free electrons on the electric permittivity (by means of a simple harmonic oscillator model). We could have incorporated the same model in other parts of space, to better model conducting metallic elements of these parts as well. For the transformation optics construction we use the approximate scheme introduced in [10], which is based on a transformation blowing up a small ball of radius ε to the cloaked region. ∗
EPFL SB MATHAA CAMA, Station 8, CH-1015 Lausanne, Switzerland, [email protected] Department of Mathematics, Rutgers University, New Brunswick, NJ 08903, USA, [email protected] †
1
When viewed in (complex) frequency domain, the refractive index associated with the Drude-Lorentz model may be extended analytically to the whole upper half plane. As is well known, an immediate consequence of this is causality for the associated non-local wave equation, see [8] and [28], – a property which is most essential for the well-posedness (and the physical relevance) of this equation. Another well known consequence of this analyticity property are the so-called Kramers-Kr¨onig relations between the real and the imaginary part of the refractive index (they are essentially related by Hilbert transforms). However, this fact is not explicitly used in our analysis. Approximate cloaking schemes for the Helmholtz equation based on the regularized transformations introduced in [10] have been studied extensively in various regimes, see [9, 16, 17, 21]. A related scheme, which (in 3d) blows up a small diameter cylinder to the cloaked region is studied in [18], under the condition that the cross section of the cylinder is symmetric, the degree of visibility of this scheme is the same as for the 3d scheme considered in [9, 16, 21], even though the material properties are less singular. The corresponding perfect cloaking scheme was introduced in [5, 13], and other approximate schemes are studied in [14], though no rates of convergence are established. Frequently a (damping) lossy layer is employed inside the transformation cloak. Without this lossy layer, the field inside the cloaked region might depend on the field outside (even for a perfect cloak), and resonance can appear and destroy the cloaking (or approximate cloaking) ability of the pure transformation cloak, see [17]. A discussion of the close connection between cloaking for the Helmholtz equations and quantum cloaking is given in [18]. We next describe the setting in details. Given r > 0, let Br denote the ball centered at 0 and of radius r. Let Fε be the standard transformation Rd → Rd , d = 2, 3, which blows up the ball Bε to B1 , equals the identity outside B2 , and is given by x if x ∈ Rd \ B2 , 2 − 2ε |x| x + if x ∈ B2 \ Bε , Fε (x) = (1.1) 2−ε 2 − ε |x| x if x ∈ Bε . ε Assume that the cloaked region is the ball B1/2 , the contents of which is characterized by a real, matrix valued function a and a complex function σ. The surrounding cloak contains two parts. In the time harmonic regime, these can be described as follows. The outer part is the Drude-Lorentz version of the standard layer, generated by the blow up map Fε . In this layer, occupying B2 \ B1 , the material characteristics are given by Fε ∗ I, (Fε ∗ 1 + σ1,c , (1.2) where σ1,c (k, x) =
kε2
σN . − k 2 − iσD k
(1.3)
While the first part (Fε ∗ 1 of the refractive index in (1.2) is standard from the transformation optics approach, the second part σ1,c is exactly the correction introduced by the 2
Drude-Lorentz model, see e.g., [8, page 331]. Here σN and σD are material constants, and kε > 0 is the so-called resonant frequency of the Drude-Lorentz model; in a more general model there could be several resonant frequencies {ki,ε }, and the corresponding part of the refractive index would be a sum of terms (1.3) ranging over all these frequencies, see e.g., [8, page 310]. In this paper, we use the standard notation F∗ A(y) =
DF (x)A(x)DF T (x) , | det DF (x)|
F∗ Σ(y) =
Σ(x) , | det DF (x)|
x = F −1 (y) ,
for the “pushforward” of a symmetric, matrix valued function, A, and a scalar function, Σ, by the diffeomorphism F . In what follows, we assume for ease of notation that σN = σD = 1 in B2 \ B1 . The inner part of our cloak is a fixed damping layer as considered in [16]. This damping (lossy) layer occupies B1 \ B1/2 , and its material characteristics are given by I, 1 +
i . k
Therefore, in the time harmonic regime, i.e., in frequency domain, the entire medium is characterized by 1 I, 1 in Rd \ B2 , Fε I, Fε 1 + σ1,c in B2 \ B1 , ∗ ∗ Ac , Σc := (1.4) I, 1 + i/k in B1 \ B1/2 , a, σ in B1/2 . We assume that a, σ ∈ L∞ (B1/2 ), with 1 2 |ξ| ≤ haξ, ξi ≤ Λ|ξ|2 , Λ
1 ≤ 0, we used Ac = I, Σc = ε2 + kεiγ , for n = 2, and Ac = εI, Σc = ε3 + iε k , for n = 3. This change is, however, not essential – the essential change is in the layer B2 \ B1 , with the inclusion of σ1,c . It would be interesting to investigate whether, in view of the damping present in σ1,c , the layer B1 \ B1/2 is necessary at all. 2 where we extend the time domain field by 0 for negative time.
3
The corresponding field intime domain (for positive time) is the unique weak solution uc ∈ L∞ (0, +∞); H 1 (Rd ) , with ∂t uc ∈ L∞ [0, +∞); L2 (Rd ) , to the non-local wave equation ( 2 u − div(A ∇u ) + Σ ∂ u + G ∗ ∂ u = f Σ1,c ∂tt in [0, +∞) × Rd , c c c 2,c t c t c (1.6) ∂t uc (t = 0) = uc (t = 0) = 0 in Rd , where f ∈ L2 (0, +∞) × Rd with compact support. The definition of weak solutions to (1.6), and the proof of well-posedness of (1.6) is presented in Section 4. The coefficients Σ1,2 and Σ2,c are given by 1 in Rd \ B2 , 0 in Rd \ B2 , (Fε )∗ 1 in B2 \ B1 , 0 in B2 \ B1 , Σ1,c = Σ2,c = 1 in B1 \ B1/2 , 1 in B1 \ B1/2 , σ in B1/2 , 0 in B1/2 , and G(t, x) is such that b x) = −ikσ1,c (k, x) x ∈ B2 \ B1 . G(k, A computation (see, e.g., [8, (7.110)]) shows that G(t, x) = φ(t)H(t) , where H(t) denotes the Heaviside function, i.e., ( 0 if t < 0 , H(t) = 1 otherwise , and
(1.7)
(1.8)
√ φ(t) =
2π −t/2 ∂t e sin(γ0 t) , γ0
(1.9)
with γ0 =
p kε2 − 1/4 .
(1.10)
We assume that kε > 1/2, so that γ0 is real and positive. The presence of the Heaviside function in the formula (1.7) implies causality and plays an important role in our analysis; in particular for the proof of well-posedness of uc , and to establish that the Fourier transform, u ˆc , satisfies the outgoing radiation condition. We only consider zero initial conditions. This is just for ease and simplicity of presentation; indeed, our method would work for the general case, using an approach similar to that in [20].
4
Given f , the corresponding field in the homogeneous medium without the cloak and ∞ (0, +∞); H 1 (Rd ) , with the cloaked region is the unique weak solution u ∈ L ∂t u ∈ L∞ (0, +∞); L2 (Rd ) , to the system ( 2 u − ∆u = f ∂tt in (0, +∞) × Rd , ∂t u(t = 0) = u(t = 0) = 0
in Rd .
The extent to which we have succeeded in hiding the contents of B1/2 and the cloak itself, should be measured in terms of the difference between uc and u, outside B2 . The main Theorem of this paper gives an estimate of this difference for the scheme in (1.4) Theorem 1. Let d = 2 or 3, and let f ∈ C ∞ ([0, +∞) × Rd ) be such that supp f ⊂ (0, R) × (BR \ B2 ) for some R > 0. Suppose c∗ ε−d/2 < kε < C∗ ε−K for some positive constant c∗ , C∗ and K > d/2. Given any integer M ≥ 2d + 4K − 2, there exists a constant C such that sup kuc − ukL2 (B5 \B2 ) ≤ CεT kf kC M ([0,R];L2 (BR ))
∀ T > 0,
for d = 3 ,
0 0, for d = 3 , 0 0. Suppose f ∈ L2 (Rd ) with supp f ⊂ B5 , and let 1 (Rd ) be the unique outgoing solution to vk ∈ Hloc ∆vk + k 2 vk = f in Rd . Then, for d = 2 and 0 ≤ k ≤ 1/2, k∇vk kL2 (B6 ) + kvk kL2 (B6 ) ≤ C| ln k|kf kL2 , and for d = 3 or for d = 2 and k > 1/2, k∇vk kL2 (B6 ) + (k + 1)kvk kL2 (B6 ) ≤ Ckf kL2 . Here C is a positive constant independent of k and f . 7
Proof. The conclusion in the case k < k0 , for arbitrary fixed k0 > 0, follows directly from the properties of the fundamental solution to the Helmholtz equation. The conclusion in the case k ≥ k0 can also be obtained from the fundamental solution to the Helmholtz equation. In this case, one can alternately obtain the conclusion using the Morawetz multipliers (see, e.g., [21, Lemma 2 and Proposition 1]). We note that the estimate in [21, Proposition 1] requires a damping layer due to the desire to obtain estimates that are independent of the arbitrary coefficients inside B1/2 . Since the operator here is ∆ + k 2 throughout, there is no need for such a layer. The details are left to the reader. We next recall the following result which will be used frequently in this paper. The result is from [16, Lemma 2.2] (see also [21, Lemma 3]). Lemma 2. Let d = 2 or 3, and let D be a smooth, open bounded subset of Rd such that Rd \D is connected. Suppose 0 < k < τ , for some fixed τ > 0, and suppose gk ∈ H 1/2 (∂D). 1 (Rd \ D) be the unique outgoing solution to Let vk ∈ Hloc ( ∆vk + k 2 vk = 0 in Rd \ D , vk = gk
on ∂D .
Then kvk kH 1 (BR \D) ≤ Cr kgk kH 1/2 (∂D)
for any R > 0 .
The constant CR is independent of k and gk . Furthermore for any ε > 0 sufficiently small that D ⊂ B2/ε kvk kL2 (B5/ε \B2/ε ) ≤ Cε−1/2 kgk kH 1/2 (∂D) if d = 3 (1)
−1 |H0 (k/ε)| kgk kH 1/2 (∂D) kvk kL2 (B5/ε \B2/ε ) ≤ Cε (1) |H0 (k)|
if
d=2.
Here the constant C is independent of k, gk and . Finally, if we assume that gk → g 1 (Rd \ D) where v ∈ W 1 (Rd \ D) weakly in H 1/2 (∂D) as k → 0, then vk → v weakly in Hloc is the unique solution of ( ∆v = 0 in Rd \ D , v=g
on ∂D .
We next establish an estimate for the model without σ1,c , for frequency at most 1/ε. Lemma 3. Let d = 2 or 3, and let a and σ be in L∞ (B1/2 ), with a real symmetric, uniformly positive definite, and =(σ) ≥ 0 .
(2.1)
Suppose 0 < ε < τ , and 0 < k < τ /ε for some fixed, positive constant τ . For g ∈ 1 1 (Rd ) be the unique outgoing solution to H − 2 (∂B1 ) let vε ∈ Hloc ¯1 , ∆vε + ε2 k 2 vε = 0 in Rd \ B div(A∇vε ) + k 2 Σvε = 0 in B1 , (2.2) 1 ∂vε ∂vε − = g on ∂B1 . ∂r ext εd−2 ∂r int 8
Here ( A=
Rd
I
if x ∈
\ B1/2 ,
a
if x ∈ B1/2 ,
Σ=
if x ∈ Rd \ B1 ,
1
if x ∈ B1 \ B1/2 ,
1 + i/k
if x ∈ B1/2 .
σ
There exists a positive constant C, depending only on d and τ , such that kvε kH 1 (B5 \B1 ) ≤ C max{k 3−d , εd−2 /k}kgk
1
H − 2 (∂B1 )
.
Proof of Lemma 3. We follow the strategy in the proof of [16, Lemma 2.4], and consider the case d = 2 and d = 3 separately. Case 1: d = 2. We first prove kvε kL2 (B5 \B1 ) ≤ C max{k, 1/k}kgk
1
H − 2 (∂B1 )
,
(2.3) 1
by contradiction. Suppose this estimate is not true. Then there exist (gn ) ⊂ H − 2 (∂B1 ), (εn ), (kn ), (an ), and (σn ) such that 0 < εn < τ , 0 < kn < τ /εn , an and σn satisfy (2.1), and kvn kL2 (B5 \B1 ) = 1 , lim max{kn , 1/kn }kgn k − 21 =0. (2.4) H
n→∞
Here vn ∈
H1
loc
(R2 )
(∂B1 )
is the unique outgoing solution to ∆vn + ε2n kn2 vn = 0 in R2 \ B1 , div(An ∇vn ) + kn2 Σn vn = 0 in B1 , ∂vn − ∂vn = gn on ∂B1 , ∂r ext ∂r int
(2.5)
where An and Σn are defined the same way as A and Σ, with a and σ replaced by an and σn . Multiplying the equation for vn by v¯n (the conjugate of vn ) and integrating on BR , we obtain Z Z Z 2 2 2 ∂r vn v¯n − |∇vn | + εn kn |vn |2 ∂BR
BR \B1
BR \B1
Z −
hAn ∇vn , ∇¯ vn i + B1
kn2
Z
2
Z
Σn |vn | = B1
gn v¯n . (2.6) ∂B1
Letting R → ∞ in (2.6), using the outgoing condition, and considering the imaginary part, we derive that Z kn |vn |2 ≤ kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) . (2.7) B1 \B1/2
9
By Caccioppoli’s inequality, it follows that Z Z 2 2 |∇vn | ≤ C(kn + 1) B4/5 \B3/5
|vn |2
B1 \B1/2
≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) . Here and in the remainder of this proof, C denotes a positive constant depending only on d and τ (which might change from one place to another). The above estimate implies that for some r ∈ (3/5, 4/5) (r depends on n), Z Z |vn |2 ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) . (2.8) |∇vn |2 + (1 + kn2 ) ∂Br
∂Br
Multiplying the equation for vn by v¯n and integrating on B5 \ Br , we have Z
Z
Z
∂r vn v¯n − ∂B5
∂r vn v¯n − B5 \Br
∂Br
|∇vn |2 + ε2n kn2
Z
+ kn2
|vn |2
B5 \B1
Z
Σn |vn |2 =
B1 \Br
Z gn v¯n . (2.9) ∂B1
1 (R2 \ B ) is the unique outgoing solution to ∆v + ε2 k 2 v = 0 in R2 \ B Since vn ∈ Hloc 3 n 3 n n n and εn kn ≤ τ , it follows that (see, e.g., Lemma 2)
kvn kH 1 (B6 \B3 ) ≤ Ckvn kH 1/2 (∂B3 ) ,
(2.10)
Since ∆vn + ε2n kn2 vn = 0 in B5 \ B1 , using the standard theory of elliptic equations, we have that kvn kH 1/2 (∂B3 ) ≤ Ckvn kH 1 (B4 \B2 ) ≤ Ckvn kL2 (B5 \B1 ) . (2.11) A combination of (4.25) and (4.26) yields kvn kH 1 (B6 \B3 ) ≤ Ckvn kL2 (B5 \B1 ) .
(2.12)
Using (2.7), (2.8), and (2.12), we derive from (2.9) that Z |∇vn |2 ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) + Ckvn k2L2 (B5 \B1 ) . (2.13) B5 \Br
We immediately obtain from (2.13) that Z B5 \B1
|∇vn |2 +
Z
|vn |2
B5 \B1
≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) + Ckvn k2L2 (B5 \B1 ) ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) + kvn kL2 (B5 \B1 ) kvn kH 1 (B5 \B1 ) , (2.14)
10
and so, by (2.4) kvn kH 1 (B5 \B1 ) ≤ C ,
and kvn kH 1/2 (∂B1 ) ≤ C .
(2.15)
From (2.7) and (2.15), we conclude Z 2 (1 + kn ) |vn |2 ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) , B1 \B1/2
so by (2.4) lim (1 +
n→∞
kn2 )
Z
|vn |2 = 0 .
(2.16)
B1 \B1/2
Since, for any v ∈ H 1 (B1 \ Br ), kvk2L2 (∂B1 ) ≤ CkvkL2 (B1 \Br ) kvkH 1 (B1 \Br ) , (see [7, Lemma 5.5]), it follows from (2.13), (2.15), and (2.16) that lim kvn kL2 (∂B1 ) = 0 .
n→∞
We have (see, e.g., Lemma 2) for any R > 1, kvn kH 1 (BR \B1 ) ≤ CR kvn kH 1/2 (∂B1 ) ≤ CR , where we used the second estimate of (2.15) to obtain the last bound. By extraction of a subsequence (and a diagonalization argument) one might assume that εn kn → ω ∈ [0, τ ] 1 (R2 \ B ), v | 1/2 (∂B ). (since εn kn ∈ [0, τ ]) and vn → v weakly in Hloc 1 n ∂B1 → 0 weakly in H 1 By (2.4), kvkL2 (B5 \B1 ) = 1 , (2.17) and for ω > 0, v is the unique outgoing solution to 3 ( ∆v + ω 2 v = 0 in R2 \ B1 v=0
on ∂B1 .
Hence v = 0, and so we have a contradiction to (2.17). If ω = 0, then by Lemma 2, v ∈ W 1 (R2 ) is the unique such solution to ( ∆v = 0 in R2 \ B1 v=0
on ∂B1 .
Hence v = 0, and so again we have a contradiction to (2.17). This verifies the L2 estimate (2.3). We have, as in (2.13), Z Z 2 |∇vε | ≤ C max{k, 1/k}kgkH −1/2 (∂B1 ) kvε kH 1/2 (∂B1 ) + C |vε |2 , B5 \B1
B5 \B1
3
The outgoing property of v is just a consequence of the fact that the fundamental solution of the Helmholz equation with frequency εn kn converges to the fundamental solution of the Helmholtz equation with frequency ω, since ω > 0.
11
and so by (2.3) kvε kH 1 (B5 \B1 ) ≤ C max{k, 1/k}kgkH −1/2 (∂B1 ) , as desired. This completes the proof of the lemma for d = 2. Case 2: d = 3. We have (see, e.g., Lemma 2) kvε kH 1 (B5 \B1 ) ≤ Ckvε kH 1/2 (∂B1 ) .
(2.18)
Hence it suffices to prove that kvε k
1
H 2 (∂B1 )
≤ C max{1, ε/k}kgk
1
H − 2 (∂B1 )
.
(2.19)
We first prove (2.19) by contradiction for ε ≤ ε0 , with ε0 sufficiently small. Suppose this 1 is not true. Then there exist (gn ) ⊂ H − 2 (∂B1 ), (εn ), (kn ), (an ), and (σn ) such that 0 < εn < τ , 0 < kn < τ /εn , an and σn satisfy (2.1), εn → 0, and kvn k
1
H 2 (∂B1 )
lim max{1, εn /kn }kgn k
=1,
n→∞
1
H − 2 (∂B1 )
=0.
(2.20)
1 (R3 ) is the unique outgoing solution to Here vn ∈ Hloc
∆vn + ε2n kn2 vn = 0 in R3 \ B1 , div(An ∇vn ) + kn2 Σn vn = 0 in B1 , ∂vn 1 ∂vn − = gn on ∂B1 , ∂r ext εn ∂r int
(2.21)
where An and Σn are defined in the same way as A and Σ, but with a and σ replaced by an and σn . Since kvn k 21 = 1, it follows from (2.18) that H (∂B1 )
kvn kH 1 (B5 \B1 ) ≤ C . In combination with (2.20), (2.21), and the fact that εn → 0 this implies
∂v
n lim =0. 1 n→∞ ∂r int H − 2 (∂B1 )
(2.22)
Multiplying the equation of vn by v¯n and integrating on BR , we obtain Z Z Z 2 2 2 ∂r vn v¯n − |∇vn | + εn kn |vn |2 ∂BR
BR \B1
BR \B1
−
1 εn
Z hAn ∇vn , ∇¯ vn i + B1
kn2 εn
Z B1
Σn |vn |2 =
Z gn v¯n . (2.23) ∂B1
Letting R → ∞ in (2.23), using the outgoing condition, and considering the imaginary part, we derive from (2.20) and the fact kn εn ≤ τ that Z 2 lim (1 + kn ) |vn |2 = 0 . (2.24) n→+∞
B1 \B1/2
12
Since ∆vn + kn2 (1 + i/kn )vn = 0 in B1 \ B1/2 , by Caccioppoli’s inequality, we obtain Z Z |∇vn |2 ≤ C(kn2 + 1) |vn |2 . (2.25) B4/5 \B3/5
B1 \B1/2
It follows from (2.24) and (2.25) that there exits r ∈ (3/5, 4/5) (r depends on n) such that Z Z 2 2 |vn |2 → 0 as n → ∞. (2.26) |∇vn | + (1 + kn ) ∂Br
∂Br
Since ∆vn + kn2 (1 + i/kn )vn = 0 in B1 \ Br , we have Z Z Z 2 2 2 |vn | = |∇vn | + (kn + ikn ) − B1 \Br
B1 \Br
Z ∂r vn v¯n −
∂r vn v¯n .
(2.27)
∂B1
∂Br
A combination of (2.20), (2.22), (2.24), (2.26), and (2.27) yields Z lim |∇vn |2 = 0 .
(2.28)
n→∞ B \B 1 4/5
From (2.24) and (2.28), we conclude that lim kvn k
n→∞
1
H 2 (∂B1 )
≤ C lim kvn kH 1 (B1 \B4/5 ) = 0 . n→∞
this is a contradiction to (2.20), and thus (2.19) holds under the additional assumption that ε ≤ ε0 for some fixed 0 < ε0 , sufficiently small. It remains to prove (2.19) for ε0 < ε < τ . In this case, we first prove that kvε kL2 (B5 \B1 ) ≤ C max{k, 1/k}kgk
1
,
(2.29)
1
.
(2.30)
H − 2 (∂B1 )
by contradiction, and then we show that kvε kH 1 (B5 \B1 ) ≤ C max{k, 1/k}kgk
H − 2 (∂B1 )
We note that since k is bounded (k < τ /ε0 ) and ε is bounded away from zero (2.30) implies (2.19). In the argument by contradiction one may without loss of generality assume the εn converge to ε1 > 0. Thus the system (2.21) is asymptotically similar to the 2d system (2.5), and so the argument of proof proceeds in the same fashion as in the two dimensional case presented above. The details are left to the reader. We next provide some useful estimates for σ1,ε which is defined as follows σ1,ε := Fε−1 ∗ σ1,c , in B2 \ Bε .
(2.31)
Lemma 4. Assume kε ≥ c∗ ε−1 , for some fixed constant c∗ > 0. We have |σ1,ε | ≤
C1 d−1 ε kε2
if k
0) is the unique outgoing solution to Then u ˆc ∈ Hloc
div(Ac ∇ˆ uc ) + k 2 Σc u ˆc = −fˆ , where (Ac , Σc ) is given in (1.4) (see Proposition 2 in Section 4). 1 (Rd ) is the unique outgoing solution to Define u ˜ε (k, x) = u ˆc (k, Fε (k, x)). Then u ˜ε ∈ Hloc
div(Aε ∇˜ uε ) + k 2 Σε u ˜ε = −fˆ, 4
After extending uc by 0 for t < 0.
14
(3.1)
where
I, 1 I, σε (x) := 1 + σ1,ε (x) Aε , Σε = 1 1 I, d (1 + i/k) d−2 ε ε 1 1 a(εx), d σ(εx) d−2 ε ε
in Rd \ B2 , in B2 \ Bε , in Bε \ Bε/2 ,
(3.2)
in Bε/2 ,
and σ1,ε = Fε−1 ∗ σ1,c .
(3.3)
In this section, we establish the stability for solutions to (3.1), (3.2) for quite general σ1,ε ; hence in the remainder of this section, we do not assume that σ1,ε is of the form (3.3), but only that it satisfies certain bounds. We recall that a is bounded, uniformly elliptic, and σ ∈ L∞ (B1 ) with =(σ) ≥ 0 .
(3.4)
The first result of this section concerns the small to moderate frequency regime. Lemma 5. Let d = 2 or 3, τ > 0, 0 < ε, k < τ , and g ∈ L2 (Rd ) with supp g ⊂ B4 \ Bε . Assume that kσ1,ε kL∞ (B2 \Bε ) ≤ C0 , and =(σ1,ε ) ≥ 0 . (3.5) 1 (Rd ) be the unique outgoing solution to Let vε ∈ Hloc
div(Aε ∇vε ) + k 2 Σε vε = g in Rd .
(3.6)
There exists a positive constant C, depending only on d, τ and C0 , such that kvε kL2 (B5 \Bε ) ≤ C max{1, 1/k}kgkL2 .
(3.7)
Remark 1. In Lemma 5, the support of g is assumed to be inside B4 \Bε not B4 \B2 , since g will be of the form −k 2 σ1,ε u ˆ1,ε , when we apply this lemma in the proof of Theorem 2. The blow up technique does not work for Lemma 5 due to the presence of σ1,ε 6= 0 inside B2 \ Bε . It is not essential that the support of g be inside B4 \ Bε , this could be replaced by BM \ Bε for any M > 4. The constant C in the estimate would depend on M . Proof. The proof is based on a contradiction argument, in which we use an argument of removable singularity. Suppose (3.7) does not hold. Then there exist {kn }, {εn } ⊂ (0, τ ), σ1,n , an , σn , and {gn }, supp gn ⊂ B4 \ Bεn , such that (3.5) holds for σ1,n , an and σn satisfy (3.4), and max{1, 1/kn }kgn kL2 → 0 as n → ∞, kvn kL2 (B5 \Bεn ) = 1 . (3.8) 1 (Rd ) is the unique outgoing solution to Here vn ∈ Hloc
div(An ∇vn ) + kn2 Σn vn = gn in Rd ,
15
where An , Σn are defined in the same way as Aε , Σε with k, ε, σ1,ε , a, and σ replaced by kn , εn , σ1,n , an , and σn . Using the outgoing radiation condition, as in (2.24), we obtain, Z Z kn 2 |gn ||vn | . |vn | ≤ εdn Bεn \Bεn /2 Rd Here we also used that =(σ1,n ) and =(σn ) are non-negative. Since supp gn ⊂ B4 \ Bεn and kvn kL2 (B5 \Bεn ) = 1, the above inequality implies that Z
kn + 1 εd−1 n
|vn |2 ≤ 2εn max{1, 1/kn }kgn kL2 .
(3.9)
Bεn \Bεn /2
We have ∆vn +
kn2 (1 + i/kn )vn = 0 in Bεn \ Bεn /2 . ε2n
It follows from Caccioppoli’s inequality that Z Z C(kn2 + 1) 2 |vn |2 , |∇vn | ≤ 2 ε B4εn /5 \B3εn /5 Bεn \Bεn /2 n and so
εn d−2 εn (kn + 1)
Z
C(kn + 1) |∇vn | ≤ εnd−1 B4εn /5 \B3εn /5 2
Z
|vn |2 .
(3.10)
Bεn \Bεn /2
In this proof, C denotes a positive constant depending only on d and τ . From (3.9) and (3.10) Z εn |∇vn |2 ≤ Cεn max{1, 1/kn }kgn kL2 . (3.11) (k + 1) εd−2 B4εn /5 \B3εn /5 n n A combination of (3.9) and (3.11) now yields Z k + 1 1 εn n 2 2 ≤ Cεn max{1, 1/kn }kgn kL2 . |v | + |∇v | n n εn kn + 1 εd−2 B4εn /5 \B3εn /5 n It follows that for some α ∈ (3εn /5, 4εn /5) (α depends on n), Z k + 1 1 εn n 2 2 |v | + |∇v | ≤ C max{1, 1/kn }kgn kL2 → 0 n n εn kn + 1 εd−2 ∂Bα n
as n → ∞ .
(3.12) Here we used (3.8) for the last convergence assertion. Multiplying the equation for vn by v¯n , and integrating on B5 \ Bα , we have Z −
|∇vn |2 + kn2
B5 \Bεn k2 + dn 1 + εn
Z
(1 + σ1,n )|vn |2 −
B5 \Bεn
1 εnd−2
Z
|∇vn |2
Bεn \Bα
Z Z Z Z i 1 |vn |2 = gn v¯n − ∂r vn v¯n + d−2 ∂r vn v¯n . (3.13) kn Bεn \Bα εn B5 ∂B5 ∂Bα 16
1 (Rd \ B 2 d Since vn ∈ Hloc 9/2 ) is an outgoing solution to ∆vn + kn vn = 0 in R \ B9/2 , we have (see, e.g., Lemma 2) kvn kH 1 (B6 \B9/2 ) ≤ Ckvn kH 1/2 (∂B9/2 ) ,
and so, by the standard theory of elliptic equations, kvn kH 1 (B6 \B9/2 ) ≤ Ckvn kL2 (B5 \Bεn ) .
(3.14)
Using (3.5), (3.8), (3.9), (3.12), and (3.14), in combination with (3.13), we now obtain Z |∇vn |2 ≤ C . (3.15) B5 \Bα
Define un ∈ H 1 (Bα ) as follows ∆un = 0 in Bα
and
un = vn on ∂Bα .
We derive from (3.8) and (3.12) that Z |∇un |2 + |un |2 → 0 as n → ∞ .
(3.16)
Bα
Indeed, set wn (x) = un (αx) for x ∈ B1 . Then ∆wn = 0 in B1 , and kwn k2H 1 (∂B1 )
≤C α
1−d
Z
2
|un | + α
3−d
Z
2
|∇un |
≤ C max{1, 1/kn }kgn kL2 ,
∂Bα
∂Bα
where we used (3.12), and the fact that 3εn /5 < α < 4εn /5 for the last estimate. It follows that Z |∇wn |2 + |wn |2 ≤ C max{1, 1/kn }kgn kL2 , B1
which in terms of un yields Z Z −d 2 2−d α |un | + α Bα
|∇un |2 ≤ C max{1, 1/kn }kgn kL2 .
Bα
The assertion (3.16) now follows from (3.8). Define ( vn in Rd \ Bα , Vn = un in Bα . We derive from (3.8), (3.9), (3.15), and (3.16) that Z |∇Vn |2 + |Vn |2 ≤ C . B5
17
It follows that (see, e.g., Lemma 2) kVn kH 1 (BR \B4 ) ≤ CR kVn kH 1/2 (∂B4 ) ≤ CR kVn kH 1 (B5 ) ≤ CR , for any R ≥ 5, and as a consequence kVn kH 1 (BR ) ≤ CR , for all R > 0. After extraction of a subsequence we may thus assume that kn → k0 ≥ 0, εn → ε0 ≥ 0, α → α0 (recall that α depends on n), σ1,εn → σ1 weakly in L2 (σ1 satisfies 1 (Rd ). (3.5)), and Vn → V weakly in Hloc Suppose k0 > 0. If ε0 = 0 then V is an outgoing solution to the equation ∆V + k02 (1 + σ1 )V = 0 in Rd \ {0} . 1 (Rd ), it follows that Since V ∈ Hloc
∆V + k02 (1 + σ1 )V = 0 in Rd . R R Therefore, V = 0, and we have a contradiction to the fact that B5 |V |2 = lim B5 \Bε |Vn |2 n = 1. Similarly, if k0 > 0 and ε0 > 0 (and thus α0 > 0), then V is an outgoing solution to ∆V + k02 (1 + σ1 )V = 0 in Rd \ Bα0 .
(3.17)
It follows from (3.16) that V = 0 in Bα0 . Hence V |Rd \Bα0 is the unique outgoing solution to (3.19) with V = 0 on ∂Bα0 , and as a consequence V = 0 in all of Rd ; we have also arrived at a contradiction. This leaves k0 = 0. We start by considering the case ε0 > 0 (and thus α0 > 0). By Lemma 2, V ∈ W 1 (Rd \ Bα0 ) is a solution to the equation ¯α . ∆V = 0 in Rd \ B 0
(3.18)
It follows from (3.16) that V = 0 in Bα0 , and thus V is the unique solution to (3.19), with V = 0 on ∂Bα0 . Hence V = 0 in Rd \ Bα0 , and as a consequence V = 0 in Rd , so we have arrived at a contradiction. Finally this leaves only the case k0 = ε0 = 0. By Lemma 2, V ∈ W 1 (Rd ) is a solution to the equation ∆V = 0 in Rd \ {0} . (3.19) Since V ∈ W 1 (Rd ), it follows that ∆V = 0 in Rd .
(3.20)
Thus V = 0 in the case d = 3, and we have arrived at a contradiction in three dimensions. In two dimensions, we can only at present conclude that V is a constant, due to (3.20). We proceed to prove that V = 0 in the case d = 2 as well. Set v˜n (x) = vn (εn x) for x ∈ B1 \ B4/5 . 18
From (3.8), (3.9), and (3.15), we have k˜ vn k2H 1/2 (∂B1 ) ≤ C
!
Z
|∇˜ vn |2 + |˜ vn |2
Z =C
B1 \B4/5
Bε \B4ε/5
! 2 |∇vn |2 + ε−2 n |vn |
≤C .
(3.21) Since limn→∞ k˜ vn kL2 (B1 \B4/5 ) = 0 by (3.8) and (3.9), it follows from (3.21) that v˜n → 0 weakly in H 1 (B1 \ B4/5 ), and thus v˜n → 0 weakly in H 1/2 (∂B1 ) .
(3.22)
1 (R2 ) be the unique outgoing solution to Let v1,n ∈ Hloc
∆v1,n + kn2 v1,n = −kn2 σ1,n vn in R2 .5 Applying Lemma 1, the regularity theory of elliptic equations, and using (3.5) and (3.8), we have 1 k∇2 v1,n kL2 (B5 ) + k∇v1,n kL2 (B5 ) + (kn + 1)kv1,n kL2 (B5 ) ≤ Ckn2 (| ln kn | + 1) . kn + 1 As a consequence of this and the fact that kn → 0, k∇v1,n kL2 (B5 ) + kv1,n kL∞ (B5 ) ≤ Ckn2 (| ln kn | + 1) .
(3.23)
By a rescaling (remember d = 2) we get k∇˜ v1,n kL2 (B5 ) + k˜ v1,n kL∞ (B5 ) ≤ Ckn2 (| ln kn | + 1) , with v˜1,n (x) = v1,n (εn x), and thus k˜ v1,n kH 1/2 (∂B1 ) ≤ Ckn2 (| ln kn | + 1) → 0 . We define wn = vn − v1,n in R2 \ Bεn ; 1 (R2 \ B ) is the unique outgoing solution to wn ∈ Hloc εn
∆wn + kn2 wn = 0 in R2 \ Bεn ,
and
wn = vn − v1,n on ∂Bεn .
Set Wn (x) = wn (εn x) for x ∈ R2 \ B1 . 1 (R2 \ B ) is the unique outgoing solution to Then Wn ∈ Hloc 1
∆Wn + kn2 ε2n Wn = 0 in R2 \ B1 , 5
and
Setting σ1,n equal to zero outside B2 \ Bεn .
19
Wn (x) = v˜n (x) − v˜1,n (x) on ∂B1 .
(3.24)
Applying Lemma 2 for Wn and using (3.22) and (3.24), we have Wn → 0 weakly in 1 (R2 \ B ), and by interior elliptic regularity estimates Hloc 1 kWn kH 1/2 (∂B2 ) → 0 as n → 0 . Applying Lemma 2 to Wn again and rescaling, we obtain kwn kL2 (B5 \B4 ) ≤
C| ln kn | kWn kH 1/2 (∂B2 ) → 0 as n → ∞ . | ln(εn kn )|
(3.25)
A combination of (3.23) and (3.25) yields that vn → 0 in L2 (B5 \B4 ); it follows that V = 0 in B5 \ B4 , and thus V = 0 in all of R2 (since we already know it must be a constant). We have a contradiction, and the proof is complete. The second result in this section deals with the moderate to high frequency regime. Lemma 6. Let d = 2 or 3, 0 < ε < 1/2, and k > k0 > 0 for some constant k0 . Suppose 1 (Rd ) be the unique outgoing solution to g ∈ L2 (Rd ) with supp g ⊂ B4 \ Bε and let vε ∈ Hloc div(Aε ∇vε ) + k 2 Σε vε = g in Rd .
(3.26)
Assume that kσ1,ε kL∞ = χ1 ,
and
=(σ1,ε ) ≥ χ2 a.e. in B2 \ Bε ,
(3.27)
for some χ1 ≥ χ2 > 0. There exist two positive constants λ and C, independent of k, ε, χ1 , χ2 , and g such that i) If kχ1 ≤ λ, then Z
|∇vε |2 + k 2 |vε |2 ≤ C(k 4 + 1)
B5 \Bε
|g|2 .
(3.28)
Rd
B5 \Bε
ii) If kχ1 > λ, then Z
Z
Z k 2 χ4 |∇vε |2 + k 2 |vε |2 ≤ C k 4 + 2 1 |g|2 . d χ2 R
(3.29)
Remark 2. As in the previous lemma, it is not essential that the support of g be inside B4 \ Bε , this could be replaced by BM \ Bε for any M > 4. The constants in the estimates would depend on M . We also note that the estimate (3.28) is stronger than the estimate (3.29), since k > k0 > 0. It thus follows immediately that Z Z k 2 χ41 2 2 2 4 |∇vε | + k |vε | ≤ C k + 2 |g|2 , d χ B5 \Bε R 2 for all k > k0 > 0. These facts shall both be used in the proof of Theorem 2. 20
Proof. The proof is inspired by [21]. To simplify notation we drop the subscript ε from vε . Multiplying (3.26) by v¯ and integrating on BR , R > 1, we obtain Z Z Z Z ∂r v v¯ − hAε ∇v, ∇¯ vi + k2 Σε |v|2 = g¯ v. ∂BR
BR
BR
BR
Letting R go to infinity, using the outgoing condition, and considering the imaginary part, we have Z Z Z Z k 2 2 2 2 |g||v| . (3.30) k lim sup |v| + d |v| ≤ |v| + k χ2 ε Bε \Bε/2 R→∞ ∂BR Rd B2 \Bε 2
Since ∆v + kε2 v + i εk2 v = 0 in Bε \ Bε/2 and k > k0 , it follows from Caccioppoli’s inequality that Z Z Ck 2 2 |∇v| ≤ 2 |v|2 . (3.31) ε B4ε/5 \B3ε/5 Bε \Bε/2 In this proof, C denotes a positive constant independent of ε, k, χ1 , χ2 , and g. It follows from (3.30) and (3.31) that Z Z Z k2 2 d−2 2 |v| ≤ Cε k |g||v| . |∇v| + 2 ε Bε \Bε/2 Rd B4ε/5 \B3ε/5 Thus there exists t ∈ (3ε/5, 4ε/5) such that Z
|∇v|2 ≤ Cεd−3 k
Z
Z |g||v|
and
Rd
∂Bt
∂Bt
|v|2 ≤
Cεd−1 k
Z |g||v| .
(3.32)
Rd
Applying [21, Lemma 2] with α = ε and R > β ≥ 5 (β is a fixed constant which will be chosen later), we have 6 1 d−1
Z β(3 − d) |v|2 |∇v| + k |v| ≤ Fβ (ε, v+ ) − Fβ (R, v) + 3 2 Bβ \Bε BR \Bβ r Z Z +C |g|(|v 0 | + |v|) + C k 2 χ1 (|v||v 0 | + |v|2 ) . (3.33)
Z
2
2
2
Rd
B2 \Bε
Here k2 Fβ (r, v) = − 2
6
Z
1 P∗ (r)|v| − 2 ∂Br 2
Z
Z 1 P∗ (r)|v | + Q0∗ (r)|v|2 2 ∂Br ∂Br Z Z 1 1 2 0 − Q∗ (r)(|v| ) + P∗ (r)|∇∂Br v|2 , 2 ∂Br 2 ∂Br 0 2
This inequality is a variant of an inequality due to Morawetz and Ludwig [15] (see also [24]).
21
with 2β d−1 P∗ (r) = 2r d−1
β and Q∗ (r) = r 1
if r > β , if 0 < r < β ,
if r > β , if 0 < r < β .
Note that Fβ (r, v) = F (r, v) := −
k2 r d−1
Z
|v|2 −
∂Br
Z r |v 0 |2 d − 1 ∂Br Z Z 1 r 2 0 − (|v| ) + |∇∂Br v|2 , (3.34) 2 ∂Br d − 1 ∂Br
2r for 0 < r < β (where F is independent of β). Since P∗ (r) = d−1 and Q∗ (r) = 1 for 0 < r < β, Z i Z h i h 2r 2 2 ∆v + k v < vr + v = < ∆v + k 2 v x · ∇v + v . d−1 d−1 Bε \Bt Bε \Bt
We have
Abstract This paper concerns approximate cloaking by mapping for a full, but scalar wave equation, when one allows for physically relevant frequency dependence of the material properties of the cloak. The paper is a natural continuation of [20], but here we employ the Drude-Lorentz model in the cloaking layer, that is otherwise constructed by an approximate blow up transformation of the type introduced in [10]. The central mathematical problem translates into the analysis of the effect of a small inhomogeneity in the context of a non-local full wave equation.
1
Introduction
Cloaking by mapping (frequently referred to as transformation optics) was introduced by Pendry, Schurig, and Smith [23] for the Maxwell system, and Leonhardt [12] in the geometric optics setting. These authors used a singular change of variables which blows up a point to a cloaked region. The exact same transformation had been used before by Greenleaf, Lassas, and Uhlmann [6] to establish non-uniqueness in the context of the Calderon problem. The singular nature of the cloaks presents various difficulties in practice as well as in theory: (1) they are hard to fabricate and (2) in certain cases the correct definition of the corresponding electromagnetic fields is not obvious. To avoid the use of singular structures, regularized schemes have been proposed in [3, 4, 10, 26, 27]. In this paper we analyse approximate cloaking for a full wave equation using transformation optics, where we incorporate the Drude-Lorentz model, see e.g., [8], in the layer constructed by transformation optics. The Drude-Lorentz model takes into account the effect of the oscillations of free electrons on the electric permittivity (by means of a simple harmonic oscillator model). We could have incorporated the same model in other parts of space, to better model conducting metallic elements of these parts as well. For the transformation optics construction we use the approximate scheme introduced in [10], which is based on a transformation blowing up a small ball of radius ε to the cloaked region. ∗
EPFL SB MATHAA CAMA, Station 8, CH-1015 Lausanne, Switzerland, [email protected] Department of Mathematics, Rutgers University, New Brunswick, NJ 08903, USA, [email protected] †
1
When viewed in (complex) frequency domain, the refractive index associated with the Drude-Lorentz model may be extended analytically to the whole upper half plane. As is well known, an immediate consequence of this is causality for the associated non-local wave equation, see [8] and [28], – a property which is most essential for the well-posedness (and the physical relevance) of this equation. Another well known consequence of this analyticity property are the so-called Kramers-Kr¨onig relations between the real and the imaginary part of the refractive index (they are essentially related by Hilbert transforms). However, this fact is not explicitly used in our analysis. Approximate cloaking schemes for the Helmholtz equation based on the regularized transformations introduced in [10] have been studied extensively in various regimes, see [9, 16, 17, 21]. A related scheme, which (in 3d) blows up a small diameter cylinder to the cloaked region is studied in [18], under the condition that the cross section of the cylinder is symmetric, the degree of visibility of this scheme is the same as for the 3d scheme considered in [9, 16, 21], even though the material properties are less singular. The corresponding perfect cloaking scheme was introduced in [5, 13], and other approximate schemes are studied in [14], though no rates of convergence are established. Frequently a (damping) lossy layer is employed inside the transformation cloak. Without this lossy layer, the field inside the cloaked region might depend on the field outside (even for a perfect cloak), and resonance can appear and destroy the cloaking (or approximate cloaking) ability of the pure transformation cloak, see [17]. A discussion of the close connection between cloaking for the Helmholtz equations and quantum cloaking is given in [18]. We next describe the setting in details. Given r > 0, let Br denote the ball centered at 0 and of radius r. Let Fε be the standard transformation Rd → Rd , d = 2, 3, which blows up the ball Bε to B1 , equals the identity outside B2 , and is given by x if x ∈ Rd \ B2 , 2 − 2ε |x| x + if x ∈ B2 \ Bε , Fε (x) = (1.1) 2−ε 2 − ε |x| x if x ∈ Bε . ε Assume that the cloaked region is the ball B1/2 , the contents of which is characterized by a real, matrix valued function a and a complex function σ. The surrounding cloak contains two parts. In the time harmonic regime, these can be described as follows. The outer part is the Drude-Lorentz version of the standard layer, generated by the blow up map Fε . In this layer, occupying B2 \ B1 , the material characteristics are given by Fε ∗ I, (Fε ∗ 1 + σ1,c , (1.2) where σ1,c (k, x) =
kε2
σN . − k 2 − iσD k
(1.3)
While the first part (Fε ∗ 1 of the refractive index in (1.2) is standard from the transformation optics approach, the second part σ1,c is exactly the correction introduced by the 2
Drude-Lorentz model, see e.g., [8, page 331]. Here σN and σD are material constants, and kε > 0 is the so-called resonant frequency of the Drude-Lorentz model; in a more general model there could be several resonant frequencies {ki,ε }, and the corresponding part of the refractive index would be a sum of terms (1.3) ranging over all these frequencies, see e.g., [8, page 310]. In this paper, we use the standard notation F∗ A(y) =
DF (x)A(x)DF T (x) , | det DF (x)|
F∗ Σ(y) =
Σ(x) , | det DF (x)|
x = F −1 (y) ,
for the “pushforward” of a symmetric, matrix valued function, A, and a scalar function, Σ, by the diffeomorphism F . In what follows, we assume for ease of notation that σN = σD = 1 in B2 \ B1 . The inner part of our cloak is a fixed damping layer as considered in [16]. This damping (lossy) layer occupies B1 \ B1/2 , and its material characteristics are given by I, 1 +
i . k
Therefore, in the time harmonic regime, i.e., in frequency domain, the entire medium is characterized by 1 I, 1 in Rd \ B2 , Fε I, Fε 1 + σ1,c in B2 \ B1 , ∗ ∗ Ac , Σc := (1.4) I, 1 + i/k in B1 \ B1/2 , a, σ in B1/2 . We assume that a, σ ∈ L∞ (B1/2 ), with 1 2 |ξ| ≤ haξ, ξi ≤ Λ|ξ|2 , Λ
1 ≤ 0, we used Ac = I, Σc = ε2 + kεiγ , for n = 2, and Ac = εI, Σc = ε3 + iε k , for n = 3. This change is, however, not essential – the essential change is in the layer B2 \ B1 , with the inclusion of σ1,c . It would be interesting to investigate whether, in view of the damping present in σ1,c , the layer B1 \ B1/2 is necessary at all. 2 where we extend the time domain field by 0 for negative time.
3
The corresponding field intime domain (for positive time) is the unique weak solution uc ∈ L∞ (0, +∞); H 1 (Rd ) , with ∂t uc ∈ L∞ [0, +∞); L2 (Rd ) , to the non-local wave equation ( 2 u − div(A ∇u ) + Σ ∂ u + G ∗ ∂ u = f Σ1,c ∂tt in [0, +∞) × Rd , c c c 2,c t c t c (1.6) ∂t uc (t = 0) = uc (t = 0) = 0 in Rd , where f ∈ L2 (0, +∞) × Rd with compact support. The definition of weak solutions to (1.6), and the proof of well-posedness of (1.6) is presented in Section 4. The coefficients Σ1,2 and Σ2,c are given by 1 in Rd \ B2 , 0 in Rd \ B2 , (Fε )∗ 1 in B2 \ B1 , 0 in B2 \ B1 , Σ1,c = Σ2,c = 1 in B1 \ B1/2 , 1 in B1 \ B1/2 , σ in B1/2 , 0 in B1/2 , and G(t, x) is such that b x) = −ikσ1,c (k, x) x ∈ B2 \ B1 . G(k, A computation (see, e.g., [8, (7.110)]) shows that G(t, x) = φ(t)H(t) , where H(t) denotes the Heaviside function, i.e., ( 0 if t < 0 , H(t) = 1 otherwise , and
(1.7)
(1.8)
√ φ(t) =
2π −t/2 ∂t e sin(γ0 t) , γ0
(1.9)
with γ0 =
p kε2 − 1/4 .
(1.10)
We assume that kε > 1/2, so that γ0 is real and positive. The presence of the Heaviside function in the formula (1.7) implies causality and plays an important role in our analysis; in particular for the proof of well-posedness of uc , and to establish that the Fourier transform, u ˆc , satisfies the outgoing radiation condition. We only consider zero initial conditions. This is just for ease and simplicity of presentation; indeed, our method would work for the general case, using an approach similar to that in [20].
4
Given f , the corresponding field in the homogeneous medium without the cloak and ∞ (0, +∞); H 1 (Rd ) , with the cloaked region is the unique weak solution u ∈ L ∂t u ∈ L∞ (0, +∞); L2 (Rd ) , to the system ( 2 u − ∆u = f ∂tt in (0, +∞) × Rd , ∂t u(t = 0) = u(t = 0) = 0
in Rd .
The extent to which we have succeeded in hiding the contents of B1/2 and the cloak itself, should be measured in terms of the difference between uc and u, outside B2 . The main Theorem of this paper gives an estimate of this difference for the scheme in (1.4) Theorem 1. Let d = 2 or 3, and let f ∈ C ∞ ([0, +∞) × Rd ) be such that supp f ⊂ (0, R) × (BR \ B2 ) for some R > 0. Suppose c∗ ε−d/2 < kε < C∗ ε−K for some positive constant c∗ , C∗ and K > d/2. Given any integer M ≥ 2d + 4K − 2, there exists a constant C such that sup kuc − ukL2 (B5 \B2 ) ≤ CεT kf kC M ([0,R];L2 (BR ))
∀ T > 0,
for d = 3 ,
0 0, for d = 3 , 0 0. Suppose f ∈ L2 (Rd ) with supp f ⊂ B5 , and let 1 (Rd ) be the unique outgoing solution to vk ∈ Hloc ∆vk + k 2 vk = f in Rd . Then, for d = 2 and 0 ≤ k ≤ 1/2, k∇vk kL2 (B6 ) + kvk kL2 (B6 ) ≤ C| ln k|kf kL2 , and for d = 3 or for d = 2 and k > 1/2, k∇vk kL2 (B6 ) + (k + 1)kvk kL2 (B6 ) ≤ Ckf kL2 . Here C is a positive constant independent of k and f . 7
Proof. The conclusion in the case k < k0 , for arbitrary fixed k0 > 0, follows directly from the properties of the fundamental solution to the Helmholtz equation. The conclusion in the case k ≥ k0 can also be obtained from the fundamental solution to the Helmholtz equation. In this case, one can alternately obtain the conclusion using the Morawetz multipliers (see, e.g., [21, Lemma 2 and Proposition 1]). We note that the estimate in [21, Proposition 1] requires a damping layer due to the desire to obtain estimates that are independent of the arbitrary coefficients inside B1/2 . Since the operator here is ∆ + k 2 throughout, there is no need for such a layer. The details are left to the reader. We next recall the following result which will be used frequently in this paper. The result is from [16, Lemma 2.2] (see also [21, Lemma 3]). Lemma 2. Let d = 2 or 3, and let D be a smooth, open bounded subset of Rd such that Rd \D is connected. Suppose 0 < k < τ , for some fixed τ > 0, and suppose gk ∈ H 1/2 (∂D). 1 (Rd \ D) be the unique outgoing solution to Let vk ∈ Hloc ( ∆vk + k 2 vk = 0 in Rd \ D , vk = gk
on ∂D .
Then kvk kH 1 (BR \D) ≤ Cr kgk kH 1/2 (∂D)
for any R > 0 .
The constant CR is independent of k and gk . Furthermore for any ε > 0 sufficiently small that D ⊂ B2/ε kvk kL2 (B5/ε \B2/ε ) ≤ Cε−1/2 kgk kH 1/2 (∂D) if d = 3 (1)
−1 |H0 (k/ε)| kgk kH 1/2 (∂D) kvk kL2 (B5/ε \B2/ε ) ≤ Cε (1) |H0 (k)|
if
d=2.
Here the constant C is independent of k, gk and . Finally, if we assume that gk → g 1 (Rd \ D) where v ∈ W 1 (Rd \ D) weakly in H 1/2 (∂D) as k → 0, then vk → v weakly in Hloc is the unique solution of ( ∆v = 0 in Rd \ D , v=g
on ∂D .
We next establish an estimate for the model without σ1,c , for frequency at most 1/ε. Lemma 3. Let d = 2 or 3, and let a and σ be in L∞ (B1/2 ), with a real symmetric, uniformly positive definite, and =(σ) ≥ 0 .
(2.1)
Suppose 0 < ε < τ , and 0 < k < τ /ε for some fixed, positive constant τ . For g ∈ 1 1 (Rd ) be the unique outgoing solution to H − 2 (∂B1 ) let vε ∈ Hloc ¯1 , ∆vε + ε2 k 2 vε = 0 in Rd \ B div(A∇vε ) + k 2 Σvε = 0 in B1 , (2.2) 1 ∂vε ∂vε − = g on ∂B1 . ∂r ext εd−2 ∂r int 8
Here ( A=
Rd
I
if x ∈
\ B1/2 ,
a
if x ∈ B1/2 ,
Σ=
if x ∈ Rd \ B1 ,
1
if x ∈ B1 \ B1/2 ,
1 + i/k
if x ∈ B1/2 .
σ
There exists a positive constant C, depending only on d and τ , such that kvε kH 1 (B5 \B1 ) ≤ C max{k 3−d , εd−2 /k}kgk
1
H − 2 (∂B1 )
.
Proof of Lemma 3. We follow the strategy in the proof of [16, Lemma 2.4], and consider the case d = 2 and d = 3 separately. Case 1: d = 2. We first prove kvε kL2 (B5 \B1 ) ≤ C max{k, 1/k}kgk
1
H − 2 (∂B1 )
,
(2.3) 1
by contradiction. Suppose this estimate is not true. Then there exist (gn ) ⊂ H − 2 (∂B1 ), (εn ), (kn ), (an ), and (σn ) such that 0 < εn < τ , 0 < kn < τ /εn , an and σn satisfy (2.1), and kvn kL2 (B5 \B1 ) = 1 , lim max{kn , 1/kn }kgn k − 21 =0. (2.4) H
n→∞
Here vn ∈
H1
loc
(R2 )
(∂B1 )
is the unique outgoing solution to ∆vn + ε2n kn2 vn = 0 in R2 \ B1 , div(An ∇vn ) + kn2 Σn vn = 0 in B1 , ∂vn − ∂vn = gn on ∂B1 , ∂r ext ∂r int
(2.5)
where An and Σn are defined the same way as A and Σ, with a and σ replaced by an and σn . Multiplying the equation for vn by v¯n (the conjugate of vn ) and integrating on BR , we obtain Z Z Z 2 2 2 ∂r vn v¯n − |∇vn | + εn kn |vn |2 ∂BR
BR \B1
BR \B1
Z −
hAn ∇vn , ∇¯ vn i + B1
kn2
Z
2
Z
Σn |vn | = B1
gn v¯n . (2.6) ∂B1
Letting R → ∞ in (2.6), using the outgoing condition, and considering the imaginary part, we derive that Z kn |vn |2 ≤ kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) . (2.7) B1 \B1/2
9
By Caccioppoli’s inequality, it follows that Z Z 2 2 |∇vn | ≤ C(kn + 1) B4/5 \B3/5
|vn |2
B1 \B1/2
≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) . Here and in the remainder of this proof, C denotes a positive constant depending only on d and τ (which might change from one place to another). The above estimate implies that for some r ∈ (3/5, 4/5) (r depends on n), Z Z |vn |2 ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) . (2.8) |∇vn |2 + (1 + kn2 ) ∂Br
∂Br
Multiplying the equation for vn by v¯n and integrating on B5 \ Br , we have Z
Z
Z
∂r vn v¯n − ∂B5
∂r vn v¯n − B5 \Br
∂Br
|∇vn |2 + ε2n kn2
Z
+ kn2
|vn |2
B5 \B1
Z
Σn |vn |2 =
B1 \Br
Z gn v¯n . (2.9) ∂B1
1 (R2 \ B ) is the unique outgoing solution to ∆v + ε2 k 2 v = 0 in R2 \ B Since vn ∈ Hloc 3 n 3 n n n and εn kn ≤ τ , it follows that (see, e.g., Lemma 2)
kvn kH 1 (B6 \B3 ) ≤ Ckvn kH 1/2 (∂B3 ) ,
(2.10)
Since ∆vn + ε2n kn2 vn = 0 in B5 \ B1 , using the standard theory of elliptic equations, we have that kvn kH 1/2 (∂B3 ) ≤ Ckvn kH 1 (B4 \B2 ) ≤ Ckvn kL2 (B5 \B1 ) . (2.11) A combination of (4.25) and (4.26) yields kvn kH 1 (B6 \B3 ) ≤ Ckvn kL2 (B5 \B1 ) .
(2.12)
Using (2.7), (2.8), and (2.12), we derive from (2.9) that Z |∇vn |2 ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) + Ckvn k2L2 (B5 \B1 ) . (2.13) B5 \Br
We immediately obtain from (2.13) that Z B5 \B1
|∇vn |2 +
Z
|vn |2
B5 \B1
≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) kvn kH 1/2 (∂B1 ) + Ckvn k2L2 (B5 \B1 ) ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) + kvn kL2 (B5 \B1 ) kvn kH 1 (B5 \B1 ) , (2.14)
10
and so, by (2.4) kvn kH 1 (B5 \B1 ) ≤ C ,
and kvn kH 1/2 (∂B1 ) ≤ C .
(2.15)
From (2.7) and (2.15), we conclude Z 2 (1 + kn ) |vn |2 ≤ C max{kn , 1/kn }kgn kH −1/2 (∂B1 ) , B1 \B1/2
so by (2.4) lim (1 +
n→∞
kn2 )
Z
|vn |2 = 0 .
(2.16)
B1 \B1/2
Since, for any v ∈ H 1 (B1 \ Br ), kvk2L2 (∂B1 ) ≤ CkvkL2 (B1 \Br ) kvkH 1 (B1 \Br ) , (see [7, Lemma 5.5]), it follows from (2.13), (2.15), and (2.16) that lim kvn kL2 (∂B1 ) = 0 .
n→∞
We have (see, e.g., Lemma 2) for any R > 1, kvn kH 1 (BR \B1 ) ≤ CR kvn kH 1/2 (∂B1 ) ≤ CR , where we used the second estimate of (2.15) to obtain the last bound. By extraction of a subsequence (and a diagonalization argument) one might assume that εn kn → ω ∈ [0, τ ] 1 (R2 \ B ), v | 1/2 (∂B ). (since εn kn ∈ [0, τ ]) and vn → v weakly in Hloc 1 n ∂B1 → 0 weakly in H 1 By (2.4), kvkL2 (B5 \B1 ) = 1 , (2.17) and for ω > 0, v is the unique outgoing solution to 3 ( ∆v + ω 2 v = 0 in R2 \ B1 v=0
on ∂B1 .
Hence v = 0, and so we have a contradiction to (2.17). If ω = 0, then by Lemma 2, v ∈ W 1 (R2 ) is the unique such solution to ( ∆v = 0 in R2 \ B1 v=0
on ∂B1 .
Hence v = 0, and so again we have a contradiction to (2.17). This verifies the L2 estimate (2.3). We have, as in (2.13), Z Z 2 |∇vε | ≤ C max{k, 1/k}kgkH −1/2 (∂B1 ) kvε kH 1/2 (∂B1 ) + C |vε |2 , B5 \B1
B5 \B1
3
The outgoing property of v is just a consequence of the fact that the fundamental solution of the Helmholz equation with frequency εn kn converges to the fundamental solution of the Helmholtz equation with frequency ω, since ω > 0.
11
and so by (2.3) kvε kH 1 (B5 \B1 ) ≤ C max{k, 1/k}kgkH −1/2 (∂B1 ) , as desired. This completes the proof of the lemma for d = 2. Case 2: d = 3. We have (see, e.g., Lemma 2) kvε kH 1 (B5 \B1 ) ≤ Ckvε kH 1/2 (∂B1 ) .
(2.18)
Hence it suffices to prove that kvε k
1
H 2 (∂B1 )
≤ C max{1, ε/k}kgk
1
H − 2 (∂B1 )
.
(2.19)
We first prove (2.19) by contradiction for ε ≤ ε0 , with ε0 sufficiently small. Suppose this 1 is not true. Then there exist (gn ) ⊂ H − 2 (∂B1 ), (εn ), (kn ), (an ), and (σn ) such that 0 < εn < τ , 0 < kn < τ /εn , an and σn satisfy (2.1), εn → 0, and kvn k
1
H 2 (∂B1 )
lim max{1, εn /kn }kgn k
=1,
n→∞
1
H − 2 (∂B1 )
=0.
(2.20)
1 (R3 ) is the unique outgoing solution to Here vn ∈ Hloc
∆vn + ε2n kn2 vn = 0 in R3 \ B1 , div(An ∇vn ) + kn2 Σn vn = 0 in B1 , ∂vn 1 ∂vn − = gn on ∂B1 , ∂r ext εn ∂r int
(2.21)
where An and Σn are defined in the same way as A and Σ, but with a and σ replaced by an and σn . Since kvn k 21 = 1, it follows from (2.18) that H (∂B1 )
kvn kH 1 (B5 \B1 ) ≤ C . In combination with (2.20), (2.21), and the fact that εn → 0 this implies
∂v
n lim =0. 1 n→∞ ∂r int H − 2 (∂B1 )
(2.22)
Multiplying the equation of vn by v¯n and integrating on BR , we obtain Z Z Z 2 2 2 ∂r vn v¯n − |∇vn | + εn kn |vn |2 ∂BR
BR \B1
BR \B1
−
1 εn
Z hAn ∇vn , ∇¯ vn i + B1
kn2 εn
Z B1
Σn |vn |2 =
Z gn v¯n . (2.23) ∂B1
Letting R → ∞ in (2.23), using the outgoing condition, and considering the imaginary part, we derive from (2.20) and the fact kn εn ≤ τ that Z 2 lim (1 + kn ) |vn |2 = 0 . (2.24) n→+∞
B1 \B1/2
12
Since ∆vn + kn2 (1 + i/kn )vn = 0 in B1 \ B1/2 , by Caccioppoli’s inequality, we obtain Z Z |∇vn |2 ≤ C(kn2 + 1) |vn |2 . (2.25) B4/5 \B3/5
B1 \B1/2
It follows from (2.24) and (2.25) that there exits r ∈ (3/5, 4/5) (r depends on n) such that Z Z 2 2 |vn |2 → 0 as n → ∞. (2.26) |∇vn | + (1 + kn ) ∂Br
∂Br
Since ∆vn + kn2 (1 + i/kn )vn = 0 in B1 \ Br , we have Z Z Z 2 2 2 |vn | = |∇vn | + (kn + ikn ) − B1 \Br
B1 \Br
Z ∂r vn v¯n −
∂r vn v¯n .
(2.27)
∂B1
∂Br
A combination of (2.20), (2.22), (2.24), (2.26), and (2.27) yields Z lim |∇vn |2 = 0 .
(2.28)
n→∞ B \B 1 4/5
From (2.24) and (2.28), we conclude that lim kvn k
n→∞
1
H 2 (∂B1 )
≤ C lim kvn kH 1 (B1 \B4/5 ) = 0 . n→∞
this is a contradiction to (2.20), and thus (2.19) holds under the additional assumption that ε ≤ ε0 for some fixed 0 < ε0 , sufficiently small. It remains to prove (2.19) for ε0 < ε < τ . In this case, we first prove that kvε kL2 (B5 \B1 ) ≤ C max{k, 1/k}kgk
1
,
(2.29)
1
.
(2.30)
H − 2 (∂B1 )
by contradiction, and then we show that kvε kH 1 (B5 \B1 ) ≤ C max{k, 1/k}kgk
H − 2 (∂B1 )
We note that since k is bounded (k < τ /ε0 ) and ε is bounded away from zero (2.30) implies (2.19). In the argument by contradiction one may without loss of generality assume the εn converge to ε1 > 0. Thus the system (2.21) is asymptotically similar to the 2d system (2.5), and so the argument of proof proceeds in the same fashion as in the two dimensional case presented above. The details are left to the reader. We next provide some useful estimates for σ1,ε which is defined as follows σ1,ε := Fε−1 ∗ σ1,c , in B2 \ Bε .
(2.31)
Lemma 4. Assume kε ≥ c∗ ε−1 , for some fixed constant c∗ > 0. We have |σ1,ε | ≤
C1 d−1 ε kε2
if k
0) is the unique outgoing solution to Then u ˆc ∈ Hloc
div(Ac ∇ˆ uc ) + k 2 Σc u ˆc = −fˆ , where (Ac , Σc ) is given in (1.4) (see Proposition 2 in Section 4). 1 (Rd ) is the unique outgoing solution to Define u ˜ε (k, x) = u ˆc (k, Fε (k, x)). Then u ˜ε ∈ Hloc
div(Aε ∇˜ uε ) + k 2 Σε u ˜ε = −fˆ, 4
After extending uc by 0 for t < 0.
14
(3.1)
where
I, 1 I, σε (x) := 1 + σ1,ε (x) Aε , Σε = 1 1 I, d (1 + i/k) d−2 ε ε 1 1 a(εx), d σ(εx) d−2 ε ε
in Rd \ B2 , in B2 \ Bε , in Bε \ Bε/2 ,
(3.2)
in Bε/2 ,
and σ1,ε = Fε−1 ∗ σ1,c .
(3.3)
In this section, we establish the stability for solutions to (3.1), (3.2) for quite general σ1,ε ; hence in the remainder of this section, we do not assume that σ1,ε is of the form (3.3), but only that it satisfies certain bounds. We recall that a is bounded, uniformly elliptic, and σ ∈ L∞ (B1 ) with =(σ) ≥ 0 .
(3.4)
The first result of this section concerns the small to moderate frequency regime. Lemma 5. Let d = 2 or 3, τ > 0, 0 < ε, k < τ , and g ∈ L2 (Rd ) with supp g ⊂ B4 \ Bε . Assume that kσ1,ε kL∞ (B2 \Bε ) ≤ C0 , and =(σ1,ε ) ≥ 0 . (3.5) 1 (Rd ) be the unique outgoing solution to Let vε ∈ Hloc
div(Aε ∇vε ) + k 2 Σε vε = g in Rd .
(3.6)
There exists a positive constant C, depending only on d, τ and C0 , such that kvε kL2 (B5 \Bε ) ≤ C max{1, 1/k}kgkL2 .
(3.7)
Remark 1. In Lemma 5, the support of g is assumed to be inside B4 \Bε not B4 \B2 , since g will be of the form −k 2 σ1,ε u ˆ1,ε , when we apply this lemma in the proof of Theorem 2. The blow up technique does not work for Lemma 5 due to the presence of σ1,ε 6= 0 inside B2 \ Bε . It is not essential that the support of g be inside B4 \ Bε , this could be replaced by BM \ Bε for any M > 4. The constant C in the estimate would depend on M . Proof. The proof is based on a contradiction argument, in which we use an argument of removable singularity. Suppose (3.7) does not hold. Then there exist {kn }, {εn } ⊂ (0, τ ), σ1,n , an , σn , and {gn }, supp gn ⊂ B4 \ Bεn , such that (3.5) holds for σ1,n , an and σn satisfy (3.4), and max{1, 1/kn }kgn kL2 → 0 as n → ∞, kvn kL2 (B5 \Bεn ) = 1 . (3.8) 1 (Rd ) is the unique outgoing solution to Here vn ∈ Hloc
div(An ∇vn ) + kn2 Σn vn = gn in Rd ,
15
where An , Σn are defined in the same way as Aε , Σε with k, ε, σ1,ε , a, and σ replaced by kn , εn , σ1,n , an , and σn . Using the outgoing radiation condition, as in (2.24), we obtain, Z Z kn 2 |gn ||vn | . |vn | ≤ εdn Bεn \Bεn /2 Rd Here we also used that =(σ1,n ) and =(σn ) are non-negative. Since supp gn ⊂ B4 \ Bεn and kvn kL2 (B5 \Bεn ) = 1, the above inequality implies that Z
kn + 1 εd−1 n
|vn |2 ≤ 2εn max{1, 1/kn }kgn kL2 .
(3.9)
Bεn \Bεn /2
We have ∆vn +
kn2 (1 + i/kn )vn = 0 in Bεn \ Bεn /2 . ε2n
It follows from Caccioppoli’s inequality that Z Z C(kn2 + 1) 2 |vn |2 , |∇vn | ≤ 2 ε B4εn /5 \B3εn /5 Bεn \Bεn /2 n and so
εn d−2 εn (kn + 1)
Z
C(kn + 1) |∇vn | ≤ εnd−1 B4εn /5 \B3εn /5 2
Z
|vn |2 .
(3.10)
Bεn \Bεn /2
In this proof, C denotes a positive constant depending only on d and τ . From (3.9) and (3.10) Z εn |∇vn |2 ≤ Cεn max{1, 1/kn }kgn kL2 . (3.11) (k + 1) εd−2 B4εn /5 \B3εn /5 n n A combination of (3.9) and (3.11) now yields Z k + 1 1 εn n 2 2 ≤ Cεn max{1, 1/kn }kgn kL2 . |v | + |∇v | n n εn kn + 1 εd−2 B4εn /5 \B3εn /5 n It follows that for some α ∈ (3εn /5, 4εn /5) (α depends on n), Z k + 1 1 εn n 2 2 |v | + |∇v | ≤ C max{1, 1/kn }kgn kL2 → 0 n n εn kn + 1 εd−2 ∂Bα n
as n → ∞ .
(3.12) Here we used (3.8) for the last convergence assertion. Multiplying the equation for vn by v¯n , and integrating on B5 \ Bα , we have Z −
|∇vn |2 + kn2
B5 \Bεn k2 + dn 1 + εn
Z
(1 + σ1,n )|vn |2 −
B5 \Bεn
1 εnd−2
Z
|∇vn |2
Bεn \Bα
Z Z Z Z i 1 |vn |2 = gn v¯n − ∂r vn v¯n + d−2 ∂r vn v¯n . (3.13) kn Bεn \Bα εn B5 ∂B5 ∂Bα 16
1 (Rd \ B 2 d Since vn ∈ Hloc 9/2 ) is an outgoing solution to ∆vn + kn vn = 0 in R \ B9/2 , we have (see, e.g., Lemma 2) kvn kH 1 (B6 \B9/2 ) ≤ Ckvn kH 1/2 (∂B9/2 ) ,
and so, by the standard theory of elliptic equations, kvn kH 1 (B6 \B9/2 ) ≤ Ckvn kL2 (B5 \Bεn ) .
(3.14)
Using (3.5), (3.8), (3.9), (3.12), and (3.14), in combination with (3.13), we now obtain Z |∇vn |2 ≤ C . (3.15) B5 \Bα
Define un ∈ H 1 (Bα ) as follows ∆un = 0 in Bα
and
un = vn on ∂Bα .
We derive from (3.8) and (3.12) that Z |∇un |2 + |un |2 → 0 as n → ∞ .
(3.16)
Bα
Indeed, set wn (x) = un (αx) for x ∈ B1 . Then ∆wn = 0 in B1 , and kwn k2H 1 (∂B1 )
≤C α
1−d
Z
2
|un | + α
3−d
Z
2
|∇un |
≤ C max{1, 1/kn }kgn kL2 ,
∂Bα
∂Bα
where we used (3.12), and the fact that 3εn /5 < α < 4εn /5 for the last estimate. It follows that Z |∇wn |2 + |wn |2 ≤ C max{1, 1/kn }kgn kL2 , B1
which in terms of un yields Z Z −d 2 2−d α |un | + α Bα
|∇un |2 ≤ C max{1, 1/kn }kgn kL2 .
Bα
The assertion (3.16) now follows from (3.8). Define ( vn in Rd \ Bα , Vn = un in Bα . We derive from (3.8), (3.9), (3.15), and (3.16) that Z |∇Vn |2 + |Vn |2 ≤ C . B5
17
It follows that (see, e.g., Lemma 2) kVn kH 1 (BR \B4 ) ≤ CR kVn kH 1/2 (∂B4 ) ≤ CR kVn kH 1 (B5 ) ≤ CR , for any R ≥ 5, and as a consequence kVn kH 1 (BR ) ≤ CR , for all R > 0. After extraction of a subsequence we may thus assume that kn → k0 ≥ 0, εn → ε0 ≥ 0, α → α0 (recall that α depends on n), σ1,εn → σ1 weakly in L2 (σ1 satisfies 1 (Rd ). (3.5)), and Vn → V weakly in Hloc Suppose k0 > 0. If ε0 = 0 then V is an outgoing solution to the equation ∆V + k02 (1 + σ1 )V = 0 in Rd \ {0} . 1 (Rd ), it follows that Since V ∈ Hloc
∆V + k02 (1 + σ1 )V = 0 in Rd . R R Therefore, V = 0, and we have a contradiction to the fact that B5 |V |2 = lim B5 \Bε |Vn |2 n = 1. Similarly, if k0 > 0 and ε0 > 0 (and thus α0 > 0), then V is an outgoing solution to ∆V + k02 (1 + σ1 )V = 0 in Rd \ Bα0 .
(3.17)
It follows from (3.16) that V = 0 in Bα0 . Hence V |Rd \Bα0 is the unique outgoing solution to (3.19) with V = 0 on ∂Bα0 , and as a consequence V = 0 in all of Rd ; we have also arrived at a contradiction. This leaves k0 = 0. We start by considering the case ε0 > 0 (and thus α0 > 0). By Lemma 2, V ∈ W 1 (Rd \ Bα0 ) is a solution to the equation ¯α . ∆V = 0 in Rd \ B 0
(3.18)
It follows from (3.16) that V = 0 in Bα0 , and thus V is the unique solution to (3.19), with V = 0 on ∂Bα0 . Hence V = 0 in Rd \ Bα0 , and as a consequence V = 0 in Rd , so we have arrived at a contradiction. Finally this leaves only the case k0 = ε0 = 0. By Lemma 2, V ∈ W 1 (Rd ) is a solution to the equation ∆V = 0 in Rd \ {0} . (3.19) Since V ∈ W 1 (Rd ), it follows that ∆V = 0 in Rd .
(3.20)
Thus V = 0 in the case d = 3, and we have arrived at a contradiction in three dimensions. In two dimensions, we can only at present conclude that V is a constant, due to (3.20). We proceed to prove that V = 0 in the case d = 2 as well. Set v˜n (x) = vn (εn x) for x ∈ B1 \ B4/5 . 18
From (3.8), (3.9), and (3.15), we have k˜ vn k2H 1/2 (∂B1 ) ≤ C
!
Z
|∇˜ vn |2 + |˜ vn |2
Z =C
B1 \B4/5
Bε \B4ε/5
! 2 |∇vn |2 + ε−2 n |vn |
≤C .
(3.21) Since limn→∞ k˜ vn kL2 (B1 \B4/5 ) = 0 by (3.8) and (3.9), it follows from (3.21) that v˜n → 0 weakly in H 1 (B1 \ B4/5 ), and thus v˜n → 0 weakly in H 1/2 (∂B1 ) .
(3.22)
1 (R2 ) be the unique outgoing solution to Let v1,n ∈ Hloc
∆v1,n + kn2 v1,n = −kn2 σ1,n vn in R2 .5 Applying Lemma 1, the regularity theory of elliptic equations, and using (3.5) and (3.8), we have 1 k∇2 v1,n kL2 (B5 ) + k∇v1,n kL2 (B5 ) + (kn + 1)kv1,n kL2 (B5 ) ≤ Ckn2 (| ln kn | + 1) . kn + 1 As a consequence of this and the fact that kn → 0, k∇v1,n kL2 (B5 ) + kv1,n kL∞ (B5 ) ≤ Ckn2 (| ln kn | + 1) .
(3.23)
By a rescaling (remember d = 2) we get k∇˜ v1,n kL2 (B5 ) + k˜ v1,n kL∞ (B5 ) ≤ Ckn2 (| ln kn | + 1) , with v˜1,n (x) = v1,n (εn x), and thus k˜ v1,n kH 1/2 (∂B1 ) ≤ Ckn2 (| ln kn | + 1) → 0 . We define wn = vn − v1,n in R2 \ Bεn ; 1 (R2 \ B ) is the unique outgoing solution to wn ∈ Hloc εn
∆wn + kn2 wn = 0 in R2 \ Bεn ,
and
wn = vn − v1,n on ∂Bεn .
Set Wn (x) = wn (εn x) for x ∈ R2 \ B1 . 1 (R2 \ B ) is the unique outgoing solution to Then Wn ∈ Hloc 1
∆Wn + kn2 ε2n Wn = 0 in R2 \ B1 , 5
and
Setting σ1,n equal to zero outside B2 \ Bεn .
19
Wn (x) = v˜n (x) − v˜1,n (x) on ∂B1 .
(3.24)
Applying Lemma 2 for Wn and using (3.22) and (3.24), we have Wn → 0 weakly in 1 (R2 \ B ), and by interior elliptic regularity estimates Hloc 1 kWn kH 1/2 (∂B2 ) → 0 as n → 0 . Applying Lemma 2 to Wn again and rescaling, we obtain kwn kL2 (B5 \B4 ) ≤
C| ln kn | kWn kH 1/2 (∂B2 ) → 0 as n → ∞ . | ln(εn kn )|
(3.25)
A combination of (3.23) and (3.25) yields that vn → 0 in L2 (B5 \B4 ); it follows that V = 0 in B5 \ B4 , and thus V = 0 in all of R2 (since we already know it must be a constant). We have a contradiction, and the proof is complete. The second result in this section deals with the moderate to high frequency regime. Lemma 6. Let d = 2 or 3, 0 < ε < 1/2, and k > k0 > 0 for some constant k0 . Suppose 1 (Rd ) be the unique outgoing solution to g ∈ L2 (Rd ) with supp g ⊂ B4 \ Bε and let vε ∈ Hloc div(Aε ∇vε ) + k 2 Σε vε = g in Rd .
(3.26)
Assume that kσ1,ε kL∞ = χ1 ,
and
=(σ1,ε ) ≥ χ2 a.e. in B2 \ Bε ,
(3.27)
for some χ1 ≥ χ2 > 0. There exist two positive constants λ and C, independent of k, ε, χ1 , χ2 , and g such that i) If kχ1 ≤ λ, then Z
|∇vε |2 + k 2 |vε |2 ≤ C(k 4 + 1)
B5 \Bε
|g|2 .
(3.28)
Rd
B5 \Bε
ii) If kχ1 > λ, then Z
Z
Z k 2 χ4 |∇vε |2 + k 2 |vε |2 ≤ C k 4 + 2 1 |g|2 . d χ2 R
(3.29)
Remark 2. As in the previous lemma, it is not essential that the support of g be inside B4 \ Bε , this could be replaced by BM \ Bε for any M > 4. The constants in the estimates would depend on M . We also note that the estimate (3.28) is stronger than the estimate (3.29), since k > k0 > 0. It thus follows immediately that Z Z k 2 χ41 2 2 2 4 |∇vε | + k |vε | ≤ C k + 2 |g|2 , d χ B5 \Bε R 2 for all k > k0 > 0. These facts shall both be used in the proof of Theorem 2. 20
Proof. The proof is inspired by [21]. To simplify notation we drop the subscript ε from vε . Multiplying (3.26) by v¯ and integrating on BR , R > 1, we obtain Z Z Z Z ∂r v v¯ − hAε ∇v, ∇¯ vi + k2 Σε |v|2 = g¯ v. ∂BR
BR
BR
BR
Letting R go to infinity, using the outgoing condition, and considering the imaginary part, we have Z Z Z Z k 2 2 2 2 |g||v| . (3.30) k lim sup |v| + d |v| ≤ |v| + k χ2 ε Bε \Bε/2 R→∞ ∂BR Rd B2 \Bε 2
Since ∆v + kε2 v + i εk2 v = 0 in Bε \ Bε/2 and k > k0 , it follows from Caccioppoli’s inequality that Z Z Ck 2 2 |∇v| ≤ 2 |v|2 . (3.31) ε B4ε/5 \B3ε/5 Bε \Bε/2 In this proof, C denotes a positive constant independent of ε, k, χ1 , χ2 , and g. It follows from (3.30) and (3.31) that Z Z Z k2 2 d−2 2 |v| ≤ Cε k |g||v| . |∇v| + 2 ε Bε \Bε/2 Rd B4ε/5 \B3ε/5 Thus there exists t ∈ (3ε/5, 4ε/5) such that Z
|∇v|2 ≤ Cεd−3 k
Z
Z |g||v|
and
Rd
∂Bt
∂Bt
|v|2 ≤
Cεd−1 k
Z |g||v| .
(3.32)
Rd
Applying [21, Lemma 2] with α = ε and R > β ≥ 5 (β is a fixed constant which will be chosen later), we have 6 1 d−1
Z β(3 − d) |v|2 |∇v| + k |v| ≤ Fβ (ε, v+ ) − Fβ (R, v) + 3 2 Bβ \Bε BR \Bβ r Z Z +C |g|(|v 0 | + |v|) + C k 2 χ1 (|v||v 0 | + |v|2 ) . (3.33)
Z
2
2
2
Rd
B2 \Bε
Here k2 Fβ (r, v) = − 2
6
Z
1 P∗ (r)|v| − 2 ∂Br 2
Z
Z 1 P∗ (r)|v | + Q0∗ (r)|v|2 2 ∂Br ∂Br Z Z 1 1 2 0 − Q∗ (r)(|v| ) + P∗ (r)|∇∂Br v|2 , 2 ∂Br 2 ∂Br 0 2
This inequality is a variant of an inequality due to Morawetz and Ludwig [15] (see also [24]).
21
with 2β d−1 P∗ (r) = 2r d−1
β and Q∗ (r) = r 1
if r > β , if 0 < r < β ,
if r > β , if 0 < r < β .
Note that Fβ (r, v) = F (r, v) := −
k2 r d−1
Z
|v|2 −
∂Br
Z r |v 0 |2 d − 1 ∂Br Z Z 1 r 2 0 − (|v| ) + |∇∂Br v|2 , (3.34) 2 ∂Br d − 1 ∂Br
2r for 0 < r < β (where F is independent of β). Since P∗ (r) = d−1 and Q∗ (r) = 1 for 0 < r < β, Z i Z h i h 2r 2 2 ∆v + k v < vr + v = < ∆v + k 2 v x · ∇v + v . d−1 d−1 Bε \Bt Bε \Bt
We have
